Math Problem Statement

Five distinct keys are to be placed in three lockers such that no locker remains empty. In how many ways can this be done?

Solution

To solve this problem, we need to determine how many ways five distinct keys can be distributed into three lockers such that no locker remains empty. This is a typical combinatorics problem involving surjections (or onto functions), where we need to assign keys to lockers while ensuring that every locker receives at least one key.

Step-by-Step Approach:

  1. Total Number of Ways to Assign Keys (Ignoring the "No Empty Locker" Condition): If there were no restriction on the number of keys in each locker, each of the five distinct keys could be placed in any of the three lockers. This gives: 35=243ways.3^5 = 243 \quad \text{ways}.

  2. Account for the "No Empty Locker" Condition: We now use the Inclusion-Exclusion Principle to subtract the cases where one or more lockers are empty.

    • Case 1: One Locker is Empty: If one locker is empty, we are left with distributing 5 keys into 2 lockers. The number of ways to do this is 252^5. However, there are 3 ways to choose which locker will be empty, so the total number of ways to assign keys where exactly one locker is empty is: 3×25=3×32=96ways.3 \times 2^5 = 3 \times 32 = 96 \quad \text{ways}.

    • Case 2: Two Lockers are Empty: If two lockers are empty, all 5 keys must go into the remaining locker. There are 33 ways to choose which two lockers will be empty, and once the two empty lockers are selected, there is only 1 way to place all keys into the remaining locker. Hence, the number of ways to assign keys where exactly two lockers are empty is: 3×1=3ways.3 \times 1 = 3 \quad \text{ways}.

  3. Apply the Inclusion-Exclusion Formula: To find the total number of valid assignments (where no locker is empty), we subtract the cases where one or more lockers are empty from the total number of ways: Total=35(3×25)+(3×1)\text{Total} = 3^5 - \left( 3 \times 2^5 \right) + \left( 3 \times 1 \right) Substituting the values: Total=24396+3=150.\text{Total} = 243 - 96 + 3 = 150.

Thus, the number of ways to distribute 5 distinct keys into 3 lockers such that no locker remains empty is 150.

Would you like more details or have any questions about the steps? Here are some related questions you might find interesting:

  1. How can we calculate the number of ways to distribute keys when some keys are identical?
  2. What is the general formula for the number of ways to assign nn distinct objects into kk boxes with no box empty?
  3. How does the inclusion-exclusion principle work for more than two conditions?
  4. What are some real-life applications of the inclusion-exclusion principle in probability and combinatorics?
  5. How can we generalize this problem to distributing nn keys into kk lockers such that no locker is empty?

Tip: The Inclusion-Exclusion Principle is a powerful tool for solving combinatorics problems, especially when there are restrictions like "no empty boxes" or "exactly one type of object per group."

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Inclusion-Exclusion Principle
Surjections

Formulas

3^5 = 243
2^5 = 32
3^5 - (3 * 2^5) + (3 * 1) = 150

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 10-12